# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Where are you most likely to be in the triangle?
From: Bill Lionheart
Date: 2017 Jan 22, 01:31 -0800

Thanks for everyone's reply. It is delightful to wake up on Sunday morning and find it is "well known to those who know" and they are on NavList of course. I was also pleased with myself for working it out (and then eventually finding what it was called).  I agree completely with Frank that we need to train to understand where the symmedian point is for bad (ie high aspect ratio) traingles. But also we should understand the limitations of the "Maximum Likelihood" estimate. There is a probability distributions on the position and as a training looking at plots of contours of probability is good to get a feeling for it - the whole "hill" not just the summit.

Just as another mathematical nuance the errors are approximately normal only after averaging a lot. For example truncation error (reading only to so many digits and neglecting the next one) is approximately uniformly distributed   If the errors were exponentally distributed (ie like exp(-|x|) rather than exp(-x^2) you would want to minimize the sum of the distances to the line. (Fermat point?)

Now what was keeping me awake, and it sounds like someone here will know, is the same problem but for a fix made using three bearings, the cocked hat of the traditional coastal navigator). Assuming the error in the bearings is normally distributed we want to find a point that minimizes the sum of squares of the angular deviations from the position lines. The angles here are taken at the objects we are taking bearings on of course rather than the vertices of the cocked hat.

Bill Lionheart

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